Q.When the numerical value of the reaction quotient (Q) is greater than the equilibrium constant (K), in which direction does the reaction proceed to reach equilibrium?

Knowing $K_C$ tells us where equilibrium lies, but a real reaction mixture is not always sitting there already. The reaction quotient, Q, is defined with exactly the same algebraic form as $K_C$ -- for $xA+yB\rightleftharpoons lC+mD$, $Q=\dfrac{[C]^l[D]^m}{[A]^x[B]^y}$ -- but evaluated using whatever concentrations happen to be present at a given instant, which need not be the equilibrium concentrations at all.

As a reaction proceeds, the concentrations of reactants and products keep changing, so Q keeps changing continuously too -- right up until the system reaches equilibrium, at which point Q becomes numerically equal to $K_C$ and (since concentrations then stop changing) stops changing itself. Comparing the current value of Q against the fixed value of $K_C$ therefore predicts, unambiguously, which direction the system still has to move:

• $Q=K_C$: the system is already at equilibrium -- no further net change.
• $Q>K_C$: too much product is present relative to equilibrium, so the reaction proceeds in reverse (toward reactants) until Q falls back down to $K_C$.
• $Q<K_C$: too little product is present relative to equilibrium, so the reaction proceeds forward (toward products) until Q rises up to $K_C$. …